The maximum area of a rectangle that can be i | vedclass

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(D) Let the length of the rectangle be $a$ and the breadth be $b$. The diagonal of the rectangle is the diameter of the circle,so $d = 2r = 2(2) = 4$.

Vedclass · Accepted Answer

$8 	ext{ sq units}$

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(D) Let the length of the rectangle be $a$ and the breadth be $b$. The diagonal of the rectangle is the diameter of the circle,so $d = 2r = 2(2) = 4$.In the right-angled triangle formed by the diagonal,$a^2 + b^2 = d^2 = 4^2 = 16$.The area of the rectangle is $A = a \cdot b$.Since $b = \sqrt{16 - a^2}$,we have $A = a \sqrt{16 - a^2}$.To maximize $A$,we maximize $A^2 = f(a) = a^2(16 - a^2) = 16a^2 - a^4$.Differentiating with respect to $a$: $f'(a) = 32a - 4a^3$.Setting $f'(a) = 0$,we get $4a(8 - a^2) = 0$. Since $a > 0$,$a^2 = 8$,so $a = 2\sqrt{2}$.Then $b = \sqrt{16 - 8} = \sqrt{8} = 2\sqrt{2}$.The maximum area is $A = (2\sqrt{2})(2\sqrt{2}) = 8 	ext{ sq units}$.

The maximum area of a rectangle that can be inscribed in a circle of radius $2 \text{ units}$ is

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